Question

The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard deviation 6.5.
(a) What are the median and the fi rst and third quartiles of the MCAT scores? What is the interquartile range?
(b) Give the interval that contains the central 80% of the MCAT scores.

Answer 1

In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of $Q_1$ such that

$\mathbb P(X$

You have

$\mathbb P(X$

For the standard normal distribution, the first quartile is about $z\approx-0.6745$, and by symmetry the third quartile would be $z\approx0.6745$. In terms of the MCAT score distribution, these values are

$\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9$
$\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7$

The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.

The central 80% of the scores have z-scores $\pm z$ such that

$\mathbb P(-z$

That leaves 10% on either side of this range, which means

$\underbrace{\mathbb P(-z$

You have

$\mathbb P(Z$

Converting to MCAT scores,

$-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0$
$1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6$

So the interval that contains the central 80% is $(17.0,33.6)$ (give or take).